Question

In the given figure, ABCD is a cyclic quadrilateral , whose diagonals intersect at $E$ such that $\angle BAC={30}^0$ , $\angle DBC={70}^0$ and AB=BC,Find $\angle ACD$.

Answer 1

$ABCD$ is a cyclic quadrilateral whose diagonal intersect at $E$.

$\angle CBD=\angle CAD$ (Angles in the same segment are equal)

$\angle CAD={70}^{\circ }$

$\angle BAD=\angle BAC+\angle CAD={30}^{\circ }+{70}^{\circ }={100}^{\circ }$

$\angle BCD+\angle BAD={180}^{\circ }$(Opposite angles of a cyclic quadrilateral)

$\angle BCD+{100}^{\circ }={180}^{\circ }$

$\angle BCD={80}^{\circ }$

In $△ABC$,

$AB=BC$ (Given)

$\angle BCA=\angle CAB$ (Angles opposite to equal sides of a triangle are equal)

$\angle BCA={30}^{\circ }$

But $\angle BCD={80}^{\circ }$.

$\angle BCA+\angle ACD={80}^{\circ }$

${30}^{\circ }+\angle ACD={80}^{\circ }$

$\angle ACD={50}^{\circ }$